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<br />* <br />The ratio of (Qp/Qb ) at X = 3.15 is found by interpolation to be 0.34, <br />max <br />indicating the peak flow at mile 12.3 to be (0.34) (95800) = 32572 cfs. <br /> <br />To determine the time to peak at the forecast point, the velocity of <br />flow at the midpoint between the dam and the forecast point must be <br />determined The ratio of the average peak flow (QX/2) to Qb in the 12.3 <br />* ma~ <br />mile reach is found on the curve at X = 1.58. <br /> <br />. <br /> <br />. <br />Xi /2 = Xi/2Xc = 12.3/2(3.9) = 1.58 <br /> <br />. <br /> <br />This ratio is approximately 0.5, indicating the midpoint peak flow to <br />be (0.5 x 95800) = 47900 cfs. This flow is multiplied by the factor (0.3 <br />+ m/10) to compute the reference flow Qref = (.3 + .82/10) 47900 = 18297. <br /> <br />Because this flow is less than Qv' Eq. (30) is used to determine the <br /> <br />reference peak depth in the reach as follows: <br /> <br />Q l/b 1/2.49 <br />h = ( ref) = (.18297) = 13.65 ft <br />ref a 27.29 <br /> <br />The midpoint reference hydraulic depth in the reach is found using <br />Eq. (32), i.e.: <br /> <br />o <br />ref <br /> <br />h <br />_ ref_ <br />- m+l- <br /> <br />13.65 <br />1.82 <br /> <br />= 7.50 <br /> <br />The average particle velocity is determined by substituting this value <br /> <br />. of Ox into Eq. (34) as follows: <br />i <br /> <br />. <br /> <br />1.49 1/2 <br />V =-5 <br />xi n <br /> <br />(Oxr3 = <br /> <br />1/2 <br />6:~Z5 (.0008) <br /> <br />2/3 <br />(7.50) <br /> <br />V = 3.48 ft/sec <br />xi <br /> <br />1-29 <br />